All the five digit numbers in which each successive digit exceeds its predecessor are arranged in the increasing order of their magnitude. The `97^(th

If the sum of the digits in a number is divisible by 9, then the number itself is divisible by 9. The least and the greatest sum of five digits...

Correct Answer - `(.^(29)C_(2)xx.^(20)C_(2))/(.^(50)C_(5))` Five tickets out of 50 can be drawn in `.^(50)C_(5)` ways. Since `x_(1) lt x_(2) lt x_(3)lt x_(4) lt x_(5)` and `x_(3) = 30`, we have `x_(1),...

Correct Answer - D Three-digit numbers are 100, 101, …, 999. Total number of such numbers is 900. The three-digit numbers (which have all same digits) are 111, 222, 333, …,...

Correct Answer - D Across the period ionic size decreases as nuclear charge increases for successive addition of an electron but down the group increases due to increase in the number...

Correct Answer - B All these elements have outermost configuration `ns^(2)np^(5)`. So they belong to Group `(12 +5 = 17)` means halogen family.

Correct Answer - C `(sqrt(F^(2)+F^(2)+2F^(2)cos theta))^(2)=3(F)(F)` `2F^(2)+2F^(2)cos theta=2F^(2)` `cos theta=(1)/(2)rArrtheta=60^(@)`

Correct Answer - D `(d)` `a_(2)=(a_(1)+a_(2))/(2)` `a_(3)=(a_(2)+a_(4))/(2)` `a_(1)=(a_(2)+a_(2012))/(2)` `a_(2012)=(a_(2001)+a_(1))/(2)` Now `a_(2)+a_(4)+…+a_(2012)=3018`…….`(i)` `2a_(2)+2a_(4)+..+2a_(2012)=6036` `:.a_(1)+a_(2)+a_(3)+a_(5)+...+a_(2011)+a_(1)=6036` `:.2(a_(1)+a_(3)+...+a_(2011))=6036` `:.a_(1)+a_(3)+...+a_(2011)=3018`.........`(ii)` By adding `(i)` and `(ii)` we get `a_(1)+a_(2)+a_(3)+...+a_(2012)=6036`

Correct Answer - C `(c )` Case I : All six digits alike i.e. `111111`, `222222`……..etc. `=5` ways Case II : `2` alike `+2` other alike. Select any three in `"^(5)C_(3)`...

Correct Answer - (a)5 (b)7 (c )13 13

Correct Answer - 19710